Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{t - 6}{t - 10} \div \dfrac{t^3 - 6t^2 - 7t}{-4t^3 + 68t^2 - 280t} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{t - 6}{t - 10} \times \dfrac{-4t^3 + 68t^2 - 280t}{t^3 - 6t^2 - 7t} $ First factor out any common factors. $a = \dfrac{t - 6}{t - 10} \times \dfrac{-4t(t^2 - 17t + 70)}{t(t^2 - 6t - 7)} $ Then factor the quadratic expressions. $a = \dfrac {t - 6} {t - 10} \times \dfrac {-4t(t - 7)(t - 10)} {t(t - 7)(t + 1)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {(t - 6) \times -4t(t - 7)(t - 10) } {(t - 10) \times t(t - 7)(t + 1) } $ $a = \dfrac {-4t(t - 7)(t - 10)(t - 6)} {t(t - 7)(t + 1)(t - 10)} $ Notice that $(t - 7)$ and $(t - 10)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-4t\cancel{(t - 7)}(t - 10)(t - 6)} {t\cancel{(t - 7)}(t + 1)(t - 10)} $ We are dividing by $t - 7$ , so $t - 7 \neq 0$ Therefore, $t \neq 7$ $a = \dfrac {-4t\cancel{(t - 7)}\cancel{(t - 10)}(t - 6)} {t\cancel{(t - 7)}(t + 1)\cancel{(t - 10)}} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $a = \dfrac {-4t(t - 6)} {t(t + 1)} $ $ a = \dfrac{-4(t - 6)}{t + 1}; t \neq 7; t \neq 10 $